Question: $f(x, y) = x^2 - x + y^3 - xy$ What are all the critical points of $f$ ? Choose 2 answers: Choose 2 answers: (Choice A) A $\left( \dfrac{3}{4}, \dfrac{1}{2} \right)$ (Choice B) B $\left( \dfrac{1}{3}, -\dfrac{1}{3} \right)$ (Choice C) C $\left( \dfrac{3 - \sqrt{5}}{8}, \dfrac{1 - \sqrt{5}}{4} \right)$ (Choice D) D $\left( \dfrac{3 + \sqrt{5}}{8}, \dfrac{1 + \sqrt{5}}{4} \right)$
A critical point of a scalar field $f$ is where $\nabla f = \bold{0}$. [What's that bolded 0?] Let's find the gradient of $f$ ! $\nabla f = \begin{bmatrix} 2x - 1 - y \\ \\ 3y^2 - x \end{bmatrix}$ We want each component of the gradient to equal zero, so we want to solve the system of equations below. $\begin{cases} 2x - 1 - y = 0 \\ \\ 3y^2 - x = 0 \end{cases}$ We can rewrite the system of equations: $\begin{cases} 6y^2 - y - 1 = 0 \\ \\ 3y^2 = x \end{cases}$ Applying the quadratic formula, the first equation has the solution $y = \dfrac{1 \pm 5}{12}$. Because $x = 3y^2$, we can find the two solutions that make $\nabla f = 0$. Therefore, $f$ has critical points at: $\begin{aligned} &\left( \dfrac{3}{4}, \dfrac{1}{2} \right) \\ \\ &\left( \dfrac{1}{3}, -\dfrac{1}{3} \right) \end{aligned}$